Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $10.5$ years; the standard deviation is $1.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living between $5.7$ and $8.9$ years.
Solution: $10.5$ $8.9$ $12.1$ $7.3$ $13.7$ $5.7$ $15.3$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $10.5$ years. We know the standard deviation is $1.6$ years, so one standard deviation below the mean is $8.9$ years and one standard deviation above the mean is $12.1$ years. Two standard deviations below the mean is $7.3$ years and two standard deviations above the mean is $13.7$ years. Three standard deviations below the mean is $5.7$ years and three standard deviations above the mean is $15.3$ years. We are interested in the probability of a meerkat living between $5.7$ and $8.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the meerkats will have lifespans within 1 standard deviation of the mean. The probability of a particular meerkat living between $5.7$ and $8.9$ years is $\color{orange}{15.85\%}$.